Question

what is the distance between (-7,-4 and (3,-8

Answer 1

Answer:

The distance between the given points is 2 √29 units.

Step-by-step-explanation:

Let the two points be A and B.

• A ≡ ( - 7, - 4 ) ≡ ( x₁, y₁ )
• B ≡ ( 3, - 8 ) ≡ ( x₂, y₂ )

We have to find the distance between these two points.

By distance formula,

d ( A, B ) = √[ ( x₁ - x₂ )² + ( y₁ - y₂ )² ]

⇒ d ( A, B ) = √{ ( - 7 - 3 )² + [ - 4 - ( - 8 ) ]² }

⇒ d ( A, B ) = √[ ( - 10 )² + ( - 4 + 8 )² ]

⇒ d ( A, B ) = √[ 100 + ( 4 )² ]

⇒ d ( A, B ) = √( 100 + 16 )

⇒ d ( A, B ) = √( 116 )

⇒ d ( A, B ) = √( 4 * 29 )

⇒ d ( A, B ) = 2 √29 units

∴ The distance between the given points is 2 √29 units.

Answer 2

SOLUTION

TO DETERMINE

The distance between (-7,- 4) and (3,-8)

FORMULA TO BE IMPLEMENTED

For the given two points $\sf{A( x_1 , y_1) \: \: and \: \: B( x_2 , y_2)}$ the distance between the points

$= \sf{ \sqrt{ {(x_2 -x_1 )}^{2} + {(y_2 -y_1 )}^{2} } }$

EVALUATION

Here the given points are (-7,- 4) and (3,-8)

Hence the required distance

$\sf = \sqrt{ {( - 7 - 3)}^{2} + {( - 4 + 8)}^{2} } \: \: \: unit$

$\sf = \sqrt{ {( - 10)}^{2} + {( 4)}^{2} } \: \: \: unit$

$\sf = \sqrt{ 100 + 16 } \: \: \: unit$

$\sf = \sqrt{ 116 } \: \: \: unit$

$\sf = \sqrt{4 \times 29 } \: \: \: unit$

$\sf = 2 \sqrt{ 29 } \: \: \: unit$

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